Invalid MySQL result resource

neridaj

Hello,

I'm getting this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in output_fns.php on line 416

Here are the two functions involved:

function get_user_info($username)
{
	// connect to db
  	$conn = db_connect();

// query user info 
$result = $conn->query("select first_name, last_name, email from user where username='$username'"); 
if (!$result)
	return false;
else
	return $result;
}

function display_folders()
{
	$userinforesult = get_user_info($username);
	if(!$userinforesult)
		echo 'nothing';
	else
		while ($row = mysql_fetch_array($result, MYSQL_NUM)) { // line 416
		    printf("Last Name: %s  Last Name: %s Email: %s", $row[0], $row[1], $row[2]);  

		}
	echo '<a href="logout.php">Logout</a>'; 
	$userfolder =  $_SESSION['valid_user'] . '/';
	echo '<table align="center" border="1" cellpadding="5"><tr><th></th></tr><tr>';
	$dir    = 'members/' . $userfolder;
	$files = scandir($dir);

foreach($files as $value) {
	if(!in_array($value, array('.', '..'))) {

		// check for folders
		if(is_dir($dir.DIRECTORY_SEPARATOR.$value)) {
    		printf('<td><a href="preview.php?pa=%s">'.
            	'<img src="'. $dir . $value .'.jpg" width=75" height="75" />'.
           		'<br />%s<a/></td>',
           		$value, $value);
		}
	}
}
echo '</tr></table>';
}

Is it because I'm not using mysql_query? I tried that and I just got connection errors so I reverted back to:
$result = $conn->query(), which at least connects and returns something. Is there an alternative to mysql_fetch_array() that I should be using instead?

Thanks for any help,

Jason

Jack_McSlay

this error means that the input value for mysql_fetch_array is something that has nothing to do with a mysql query or that it comes from a query resulting in an error.
use mysql_error() to find out what is the exact error

laserlight

This warning:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in output_fns.php on line 416

Is due to one or more of these:

Failed to connect to database server.
Failed to select database.
Failed to execute corresponding SQL statement.

Looking at the display_folders() function, it is clear that $result is an uninitialised local variable. As such, it has no corresponding SQL statement, so effectively you are trying to retrieve a result row on something that is not even a result set.

I would do something like this instead:

function display_folders()
{
    $conn = db_connect();
    $result = $conn->query("select first_name, last_name, email from user where username='$username'"); 
    if(result) {
        while ($row = mysql_fetch_array($result, MYSQL_NUM)) { // line 416
            printf("Last Name: %s  Last Name: %s Email: %s", $row[0], $row[1], $row[2]);  

        }
        // ...
    } else {
        echo 'Nothing'; // or rather, more graceful error handling.
    }
}

Of course, calling db_connect() from within a function may not be the best idea. You may want to pass it to the function instead, or maybe use a singleton (but I prefer passing).

neridaj

I can't just return the result from get_user_info()?

function get_user_info($username)
{
	// connect to db
  	$conn = db_connect();

// query user info 
$result = $conn->query("select first_name, last_name, email from user where username='$username'"); 
if (!$result)
	return false;
else
	return $result;
}

laserlight

I can't just return the result from get_user_info()?

Yes, but it could be simplified to:

function get_user_info($username)
{
	// connect to db
	$conn = db_connect();

// query user info 
return $conn->query("select first_name, last_name, email from user where username='$username'"); 
}

At that point, if you take my advice about passing the connection object, the function becomes so trivial that it could just be integrated into the caller, since it is a specialised function anyway.